Category:Non-Automonous Linear Differential Equations

Following the notes of Krešimir Josić and Robert Rosenbaum on UNSTABLE SOLUTIONS OF NON-AUTONOMOUS LINEAR DIFFERENTIAL EQUATIONS. Look for families of matrices with constant eigenvalues, with negative real parts, which result in unstable differential equations.

Introduction
The study of linear, autonomous differential equations $$x'(t)=Ax(t)$$ lead to a link between the eigenvalues of matrix A and the behavior of the solutions. However, for linear, non-autonomous equations

$$\frac{dx}{dt}=A(t)x(t)$$

it is no longer possible to use the eigenvalues of matrix A(t) characterize the stability of the system in generic way. Such an example is

$$A(t)=\begin{bmatrix} -1-9cos^{2}(6t)+12sin(6t)cos(6t) & 12cos^{2}(6t)+9sin(6t)cos(6t)\\ -12sin^{2}(6t)+9sin(6t)cos(6t) & -1-9sin^{2}(6t)-12sin(6t)cos(6t) \end{bmatrix}$$

where the eigenvalues are negative (-1 and -10 and constant with time), yet the solution is exponentially growing.



Frozen Coefficient Equations
A look into the behavior of dynamical systems characterized by matrix A(t), with negative eigenvalues, has so far been shown not to imply stability. This phenomenon contradicts what has been determined for autonomous systems. For such to be possible the norm of the solution must increase over time

$$\frac{d\left \| x(t)) \right \|^{2}}{dt}=2x'(t).x(t)=2\left [ A(t)x(t) \right ].x(t)> 0$$

for at least some values of t.

Now lets consider the autonomous, frozen coefficient system obtained by fixing t, to $$t_{0} > 0$$, $$x'(t)=A(t_{0})x(t) > 0$$, such that the solutions are evolving further away from the origin in a given time interval. Characterize the class of 2x2 matrices

$$B=\left \{ D\ with\ negative\ Re\left [ eigenvalues \right ] |\ x.Dx > 0 for\ some\ x \in \mathbb{R}^{2}\right \}$$

A solution $$x(t)$$ to a system described by A(t) can only be unstable if $$A(t) \in B$$ for some t. However if $$x(t).A(t)x(t)\leqslant 0$$ for any t and x, then $$\left \| x(t) \right \|$$ cannot increase.

Real Eigenvalues
Let us consider a matrix D with eigenvalues $$\lambda _{1},\lambda _{2}\in\mathbb{R}$$ such that $$\lambda _{1} < \lambda _{2} \leq 0$$. Furthermore, as we are only interested on the general behavior of the system, it is possible to rotate the coordinates, forcing the eigenvector corresponding to $$\lambda _{1}$$ to be on the $$x_{1}(t)$$ axis.

$$D=\begin{bmatrix} \lambda _{1} & (\lambda _{2}-\lambda _{1})cot(\delta )\\ 0 & \lambda _{2} \end{bmatrix}$$

Where $$\delta$$ is the angle between the two eigenvectors and constrained to $$0<\delta <\pi$$. Hence it is now possible to infer

$$r(x)=x.Dx=\lambda _{1}x_{1}^{2}+\lambda _{2}x_{2}^{2}+(\lambda _{2}-\lambda _{1})x_{1}x_{2}cot(\delta )$$

As we only need to determine whether r(x) is positive, it suffices to determine its maxima on the unit circle. So for the existence of a region in which $$x.Dx>0$$ the following relation is necessary

$$sin(\delta )<\frac{1-p}{1+p}$$

where $$p=\lambda _{1}/\lambda _{2}$$, noting $$0\leq p<1$$.

Complex Eigenvalues
Any 2x2 matrix D with complex eigenvalues $$\lambda _{1,2}=k\pm\sigma i$$ and eigen vector $$\begin{bmatrix} 1\\ a\pm bi\end{bmatrix}$$ can be written in the form

$$D=\begin{bmatrix} 1 & 1\\ a+bi & a-bi \end{bmatrix} \begin{bmatrix} k+\sigma i & 0\\ 0 & k-\sigma i \end{bmatrix} \begin{bmatrix} 1 & 1\\ a+bi & a-bi \end{bmatrix}^{-1}$$

where the parameter $$b>0$$. As before consider eigenvalues with negative real part, $$k<0$$, and rotate coordinates such that $$a=0$$. Now in a more convenient form

$$D=\begin{bmatrix} 1 & 1\\ bi & -bi \end{bmatrix} \begin{bmatrix} k+\sigma i & 0\\ 0 & k-\sigma i \end{bmatrix} \begin{bmatrix} 1 & 1\\ bi & -bi \end{bmatrix}^{-1}= \begin{bmatrix} k & \frac{\sigma }{b}\\ -\sigma b& k \end{bmatrix}$$

So the solutions to $$x'=Dx$$, given initial conditions $$(x_{0},y_{0})$$ are of the form

$$\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}= \begin{bmatrix} e^{kt}(x_{0}bcos(\sigma t)+y_{0}sin(\sigma t))\\ e^{kt}(y_{0}cos(\sigma t)-x_{0}bsin(\sigma t)) \end{bmatrix}$$

As before, by finding the maximum of the norm in unitary polar coordinates, it is possible to infer that $$D \in B$$ iff

$$\left | \frac{\sigma }{k} \right |>\left | \frac{2b}{2b^{2}-1} \right |$$

Stability
Want to describe equations with unstable solutions. This can be done by choosing matrices from class B and rotating them with a constant angular velocity. So let $$D \in B $$ and

$$R(t,\omega )=e^{tG(\omega )}=\begin{bmatrix} cos(\omega t) & -sin(\omega t)\\ sin(\omega t) & cos(\omega t) \end{bmatrix}$$

Henceforth lets consider matrix A(t) as $$A(t)=R(t, \omega )DR^{-1}(t, \omega)$$. Therefore the system under analysis is

$$\frac{dx}{dt}=A(t)x=(R(t, \omega )DR^{-1}(t, \omega))x$$

and a better suited coordinates are the rotating coordinates $$ y = R^{-1}(t, \omega)x$$. This system, y', has two contributions. One from the original vector field which is now autonomous $$A(0)y:=Dy$$, the other from the rotation of the plane, $$-G( \omega )y$$. So the solution becomes

$$y(t)=e^{(D-G(\omega ))t}y(0)$$ $$x(t)=R(t, \omega )e^{(D-G(\omega ))t}x(0)$$

The main point here is that for the non-autonomous system to be unstable the autonomous system in the new coordinates needs to be unstable too.

Existence of Unstable Solutions
So far given a matrix D, need to find a matrix $$G(\omega )$$ such that $$B-G(\omega )$$ has positive eigenvalues, i.e.

$$(B-G(\omega ))x= \lambda x$$ for $$\lambda x > 0$$

This can be done by considering an interval, given $$D \in B$$, of angular frequencies which correspond to unstable equations. Such an interval is

$$I=\left [ C-\sqrt{C^{2}-\lambda _{1}\lambda _{2}}, C+\sqrt{C^{2}-\lambda _{1}\lambda _{2}}\right ]$$

where

$$C=\frac{v_{1}.v_{2}}{2det(\left [ v_{1}\ v_{2} \right ])}(\lambda _{1}-\lambda _{2})$$

and the v's denote the corresponding eigenvectors.